Question

b) An object 2 cm high produces a real image 2.5 cm high ,when placed at a
distance of 15 cm from a concave mirror. Find
(i) the position of image
(ii) focal length of the concave mirror
rapid answer ​

Answer 1

Answer:

Here, (i)=The object is 19cm behind the mirror and is virtual and erect.

(ii)=Focal length of the mirror is 8.38cm

Explanation:

Here, As per our given question,

=Height of the object(ho)=2cm

=Height of the image(hi)=2.5cm

=Distance of object from the mirror(U)=15cm

=Distance of Image from the mirror(V)=?

Now, Magnification(M)=hi/ho

=M=2.5/2

=M=1.25

Now, As it is considered that,

=M=(-V)/U

=(-V)=M×U

=(-V)=1.25×(-15)

=(-V)=(-18.75)cm

=V=18.75)cm

=V=19 (By rounding off)

=V=19cm

So,Image is 19cm behind the mirror and is Virtual and erect.

Now, Focal length of the mirror(f)= 1/f=(1/V)+(1/U) -(By mirror formula)

=1/f=(1/19)+(1/15)

On taking Lcm and solving it, we get,

=1/f=(15+19)/285

=1/f=34/285

By doing reciprocal on both sides, we get,

=f=285/34

=f=8.38cm

So, Focal length of the mirror is 8.38cm.

Thank you.